Trigonometry

1. Revision

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CAST diagram: problem solving to find coordinates

Adapted from DBE Nov 2015 Grade 11, P2, Q5.1
Maths formulas

In the diagram below, $D(-24\, ; 7)$ is a point such that ${\color{red}{OD}} = 25$ and $E\hat{O}D = \theta$ , where $\theta$ is an obtuse angle.

NOTE: The following diagram is not drawn to scale.

$F$ is a point on $OD$ such that $OF = \text{2}$ . Determine the coordinates of $F$ without using a calculator.

INSTRUCTION: Type brackets around the

x

$x$ - and

y

$y$ -coordinates, and type a semicolon in between, like this: (x ; y)

Answer: Coordinates of

F

$F$ :

coordinate

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CAST diagram: problem solving to find coordinates

Adapted from DBE Nov 2015 Grade 11, P2, Q5.1
Maths formulas

In the diagram below, $P(-20\, ; -21)$ is a point such that ${\color{red}{OP}} = 29$ and $R\hat{O}P = \alpha$ , where $\alpha$ is a reflex angle.

NOTE: The following diagram is not drawn to scale.

$T$ is a point on $OP$ such that $OT = \text{3}$ . Determine the coordinates of $T$ without using a calculator.

INSTRUCTION: Type brackets around the

x

$x$ - and

y

$y$ -coordinates, and type a semicolon in between, like this: (x ; y)

Answer: Coordinates of

T

$T$ :

coordinate

5. Area, sine, and cosine rules

5. The area rule

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The area rule: finding mistakes

One of your classmates is working on a trigonometry worksheet about the area rule. The worksheet has a triangle with sides $\text{4}$ , $\text{6}$ , and $\text{7}$ , and one angle labelled $\text{59}°$ . He wrote the area rule equation below the triangle diagram. But they made a mistake and the equation is incorrect.

Incorrect equation:

\begin{matrix} A = \frac{1}{2} (6) (7) \sin 59 ° \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} \left( 6 \right) \left( 7 \right) \sin 59° \end{array}$

What should your classmate do to correct the equation?

Answer:

To make the equation correct, he should .

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The area rule: finding mistakes

One of your classmates is working on a trigonometry worksheet about the area rule. The worksheet has a triangle with sides $\text{7}$ , $\text{8}$ , and $\text{9}$ , and one angle labelled $\text{60}°$ . She wrote the area rule equation below the triangle diagram. But they made a mistake and the equation is incorrect.

Incorrect equation:

\begin{matrix} A = \frac{1}{2} (9) (8) \sin 60 ° \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} \left( 9 \right) \left( 8 \right) \sin 60° \end{array}$

What should your classmate do to correct the equation?

Answer:

To make the equation correct, she should .

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Area formulas: old and new

The area rule for a triangle is:

\begin{matrix} A = \frac{1}{2} a b \sin θ \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} a b \sin \theta \end{array}$

You should also recognise this formula for the area of a triangle:

\begin{matrix} A = \frac{1}{2} b h \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} b h \end{array}$

The two statements below are about the relationship between these equations. Complete the statements by selecting the best choice for each statement.

Answer:

For the area rule, the angle between $a$ and $b$ $\theta$ , but the angle between $b$ and $h$ must be perpendicular.
If we substitute $\theta = 90°$ into the area rule it becomes the same as .

ID is: 3555 Seed is: 1414

Area formulas: old and new

The area rule for a triangle is:

\begin{matrix} A = \frac{1}{2} a b \sin θ \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} a b \sin \theta \end{array}$

You should also recognise this formula for the area of a triangle:

\begin{matrix} A = \frac{1}{2} b h \end{matrix}

$\begin{array}{c} A = \dfrac{1}{2} b h \end{array}$

The two statements below are about the relationship between these equations. Complete the statements by selecting the best choice for each statement.

Answer:

For the area rule, the angle between $a$ and $b$ $\theta$ , but the angle between $b$ and $h$ must be perpendicular.
If we substitute $\theta = 90°$ into the area rule it becomes the same as .

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ID is: 3527 Seed is: 8208

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The flexibility of the area rule

The diagram below shows a triangle with the angles $\hat{x}$ , $\hat{y}$ , and $\hat{z}$ . The lengths of the sides are $\text{9}$ , $\text{7}$ , and $\text{6}$ units. Answer the two questions below about this triangle.

For this triangle, we can enter the area rule in three different ways. Here are two of these:

\begin{aligned} A & = \frac{1}{2} (9) (7) \sin \hat{y} \\ A & = \frac{1}{2} (7) (6) \sin \hat{z} \end{aligned}

$\begin{aligned} A & = \dfrac{1}{2}\left(9\right)\left(7\right) \sin \hat{ y } \\ A & = \dfrac{1}{2}\left(7\right)\left(6\right) \sin \hat{ z } \end{aligned}$

Which of the choices below shows the third correct way we can write the area rule for this triangle?

A	$A = \dfrac{1}{2} \left(6\right)\left(9\right) \sin \hat{ z }$
B	$A = \dfrac{1}{2}\left(6\right)\left(9\right) \sin \hat{ x }$
C	$A = \dfrac{1}{2} \left(7\right)\left(6\right) \sin \hat{ x }$
D	$A = \dfrac{1}{2} \left(6\right)\left(7\right) \sin \hat{ y }$

Answer: The third option for the area rule is choice .

Suppose we find out that $\hat{z} = 89°$ and that $\hat{y} = 41°$ . Use this information and the equations from Question $\text{1}$ to calculate the area of the triangle.

INSTRUCTION: Round your answer to $\text{2}$ decimal places.

Answer: The area is square units.
one-of
type(numeric.abserror(0.01))

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The flexibility of the area rule

The diagram below shows a triangle with the angles $\hat{y}$ , $\hat{x}$ , and $\hat{z}$ . The lengths of the sides are $\text{10}$ , $\text{12}$ , and $\text{8}$ units. Answer the two questions below about this triangle.

For this triangle, we can enter the area rule in three different ways. Here are two of these:

\begin{aligned} A & = \frac{1}{2} (8) (10) \sin \hat{y} \\ A & = \frac{1}{2} (10) (12) \sin \hat{x} \end{aligned}

$\begin{aligned} A & = \dfrac{1}{2}\left(8\right)\left(10\right) \sin \hat{ y } \\ A & = \dfrac{1}{2}\left(10\right)\left(12\right) \sin \hat{ x } \end{aligned}$

Which of the choices below shows the third correct way we can write the area rule for this triangle?

A	$A = \dfrac{1}{2} \left(8\right)\left(10\right) \sin \hat{ z }$
B	$A = \dfrac{1}{2} \left(8\right)\left(10\right) \sin \hat{ x }$
C	$A = \dfrac{1}{2}\left(12\right)\left(8\right) \sin \hat{ z }$
D	$A = \dfrac{1}{2} \left(12\right)\left(8\right) \sin \hat{ y }$

Answer: The third option for the area rule is choice .

Suppose we find out that $\hat{y} = 82°$ and that $\hat{x} = 41°$ . Use this information and the equations from Question $\text{1}$ to calculate the area of the triangle.

INSTRUCTION: Round your answer to $\text{2}$ decimal places.

Answer: The area is square units.
one-of
type(numeric.abserror(0.01))

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Area rule to the rescue

Consider a triangle $ZXY$ , shown below. Two angles are given: $\angle Z = \text{65,7}°$ and $\angle Y = \text{57,7}°$ . The sides of the triangle are $\overline{ZX} = \text{5,6}\ \text{cm}$ , $\overline{YZ} = 2 d + 6$ , and $\overline{XY} = 6 d + 2$ , where $d \in \mathbb{R}$ . The triangle may or may not be drawn to scale.

Determine the length of side $\overline{XY}$ . Then select the correct units for the length of the side.

INSTRUCTION:

Round your answer to $\text{2}$ decimal places.
If there is no answer, type No solution in the answer box and select No units for the units.

Answer:

\bar{X Y} =

$\overline{XY} =$

one-of

type(numeric.abserror(0.004))

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Area rule to the rescue

Consider a triangle $XZY$ , shown below. Two angles are given: $\angle X = \text{69,1}°$ and $\angle Y = \text{66,4}°$ . The sides of the triangle are $\overline{XZ} = \text{6,3}\ \text{mm}$ , $\overline{YX} = 5 d + 9$ , and $\overline{ZY} = 6 d + 7$ , where $d \in \mathbb{R}$ . The triangle may or may not be drawn to scale.

Determine the length of side $\overline{ZY}$ . Then select the correct units for the length of the side.

INSTRUCTION:

Round your answer to $\text{2}$ decimal places.
If there is no answer, type No solution in the answer box and select No units for the units.

Answer:

\bar{Z Y} =

$\overline{ZY} =$

one-of

type(numeric.abserror(0.004))

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Area rule calculations

An isosceles triangle, $\Delta ABC$ , has equal sides $\overline{ AB }$ and $\overline{ AC }$ . They are both $\text{19} \ \text{mm}$ long. The third side is $\text{18} \ \text{mm}$ long. The angle at $C$ is $\text{62}\text{°}$ . The figure below shows this triangle. The figure is drawn to scale.

What is the triangle's area?

INSTRUCTION:

Round your answer to two decimal places if necessary.
Do not type units with your answer (type only the number).

Answer: The area of the triangle is

mm^{2}

$\text{mm$^{2}$}$ .

numeric

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Area in terms of a variable

The triangle below is drawn to scale. It has sides with lengths $\text{5}$ , $\text{10}$ , and an unknown length, $9 f - 3$ . Two angles are given, $\hat{A} = 24°$ and $\hat{C} = 23°$ .

Which of the following is a correct expression for the area of this triangle in terms of $f$ with all numeric values rounded to two decimal places?

A	$A = \text{17,58}f -\text{5,86}$
B	$A = \text{13,73}f -\text{9,57}$
C	$A = \text{21,83}f -\text{2,59}$

Answer: The correct expression is choice .

Determine the value of $f$ .

INSTRUCTION: Use the rounded answer from Question 1. Round your answer to two decimal places.

Answer: $f =$
one-of
type(numeric.abserror(0.004))

ID is: 3565 Seed is: 3194

Area in terms of a variable

The triangle below is drawn to scale. It has sides with lengths $\text{7}$ , $\text{10}$ , and an unknown length, $8 a - 2$ . Two angles are given, $\hat{B} = 32°$ and $\hat{C} = 46°$ .

Which of the following is a correct expression for the area of this triangle in terms of $a$ with all numeric values rounded to two decimal places?

A	$A = \text{32,77}a -\text{4,07}$
B	$A = \text{28,77}a -\text{7,19}$
C	$A = \text{25,27}a -\text{10,35}$

Answer: The correct expression is choice .

Determine the value of $a$ .

INSTRUCTION: Use the rounded answer from Question 1. Round your answer to two decimal places.

Answer: $a =$
one-of
type(numeric.abserror(0.004))

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Finding area pieces

Triangle $\Delta ABC$ is inscribed in a circle. This means the vertices of the triangle lie on the circle's circumference. The radius of the circle is $\text{4,2}\ \text{m}$ . The sides of the triangle are $\overline{AC} = \text{6}\ \text{m}$ , $\overline{CB} = \text{8,3}\ \text{m}$ , and $\overline{BA} = \text{5,1}\ \text{m}$ . $\hat{B} = 46°$ and $\hat{C} = 38°$ .

Determine the area of the shaded region, the area inside the circle but outside of the triangle. Then select the correct units for the answer.

INSTRUCTION: Round your answer to three decimal places.

Answer: The shaded area is .

one-of

type(numeric.abserror(0.0004))

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Finding area pieces

Triangle $\Delta ABC$ is inscribed in a circle. This means the vertices of the triangle lie on the circle's circumference. The radius of the circle is $\text{3,7}\ \text{m}$ . The sides of the triangle are $\overline{AB} = \text{6}\ \text{m}$ , $\overline{BC} = \text{7,3}\ \text{m}$ , and $\overline{CA} = \text{5,0}\ \text{m}$ . $\hat{C} = 55°$ and $\hat{B} = 43°$ .

Determine the area of the shaded region, the area inside the circle but outside of the triangle. Then select the correct units for the answer.

INSTRUCTION: Round your answer to three decimal places.

Answer: The shaded area is .

one-of

type(numeric.abserror(0.0004))

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Working with triangles

Two triangles share a common side, as shown below. The triangles have a common side, $\overline{JK} =$ $\text{5,95}\ \text{units}$ . The sides around the perimeter have the lengths $\overline{JG} =$ $\text{5,23}\ \text{units}$ , $\overline{GK} =$ $\text{4,44}\ \text{units}$ , $\overline{KH} =$ $\text{5,85}\ \text{units}$ , and $\overline{HJ} =$ $\text{4,96}\ \text{units}$ . There are two angles given: $\angle HKJ = \text{49,7}°$ and $\angle JGK = \text{75,3}°$ . Note that the diagram is not drawn to scale.

Determine the area of $\Delta JGK$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: The area of $\Delta JGK$ is square units.
numeric
If it now given that $\angle HJK = \text{64,1}°$ , determine the angle $\angle JHK$ .

Answer: $\angle JHK =$ °
numeric

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Working with triangles

Two triangles share a common side, as shown below. The triangles have a common side, $\overline{KJ} =$ $\text{6,88}\ \text{units}$ . The sides around the perimeter have the lengths $\overline{KH} =$ $\text{4,74}\ \text{units}$ , $\overline{HJ} =$ $\text{6,43}\ \text{units}$ , $\overline{JG} =$ $\text{4,8}\ \text{units}$ , and $\overline{GK} =$ $\text{4,55}\ \text{units}$ . There are two angles given: $\angle KGJ = \text{94,5}°$ and $\angle KJH = \text{41,5}°$ . Note that the diagram is not drawn to scale.

Determine the area of $\Delta KGJ$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: The area of $\Delta KGJ$ is square units.
numeric
If it now given that $\angle JKH = \text{64}°$ , determine the angle $\angle KHJ$ .

Answer: $\angle KHJ =$ °
numeric

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Find the triangle's area

Suppose that a triangle $XYZ$ includes angle $\hat{X} = 134°$ . The side of the triangle across from this angle is $\overline{YZ} = \text{3,08}\ \text{cm}$ long. The sides adjacent to this angle are $\overline{ZX} = \text{5,29}\ \text{cm}$ long and $\overline{XY} = 25 g^{2} - 36$ . As a function of $g$ , the area of the triangle is:

\begin{matrix} A (g) = 5 g + 6 \end{matrix}

$\begin{array}{c}A(g) = 5 g + 6\end{array}$

Calculate the numerical value of the triangle's area.

INSTRUCTION: Round your answer to two decimal places.

Answer: The area is $\text{cm$^{2}$}$ .
one-of
type(numeric.abserror(0.004))
In Question $\text{1}$ , the area function for the triangle is linear, which means the area is related to $g$ as shown in this graph:

This shows that as $g$ increases, the area increases. Why does this happen?

Answer: When $g$ increases, .

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Find the triangle's area

Suppose that a triangle $ZXY$ includes angle $\hat{Z} = 53°$ . The side of the triangle across from this angle is $\overline{XY} = \text{2,55}\ \text{mm}$ long. The sides adjacent to this angle are $\overline{YZ} = \text{2,38}\ \text{mm}$ long and $\overline{ZX} = 6 g + 5$ . As a function of $g$ , the area of the triangle is:

\begin{matrix} A (g) = 42 g^{2} + 35 g \end{matrix}

$\begin{array}{c}A(g) = 42 g^{2} + 35 g\end{array}$

Calculate the numerical value of the triangle's area.

INSTRUCTION: Round your answer to two decimal places.

Answer: The area is $\text{mm$^{2}$}$ .
one-of
type(numeric.abserror(0.004))
In Question $\text{1}$ , the area function for the triangle is quadratic, which means the area is related to $g$ as shown in this graph:

This shows that as $g$ increases, the area increases. Why does this happen?

Answer: When $g$ increases, .

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6. The sine rule

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Using the sine rule

Calculate the length of the unknown side in the triangle below. One side of the triangle has a length of $\text{11,5}$ . Two angles are labelled, with values of $73°$ and $48°$ . The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The side you need is shorter than the given side

11,5

$\text{11,5}$ so check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The length of the missing side is units.

numeric

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Algebra meets the sine rule

In triangle $EFD$ below, two angles are given: $\hat{E} = 71°$ and $\hat{D} = 68°$ . Side $\overline{DE}$ and $\overline{FD}$ have lengths $9 y + 7$ $\text{cm}$ and $8 y + 2$ $\text{cm}$ respectively, where $y \in \mathbb{R}$ . The third side, $\overline{EF}$ , is $\text{20}\ \text{cm}$ long. The figure may or may not be drawn to scale.

Determine the value of $y$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: $y =$
one-of
type(numeric.abserror(0.004))
Hence determine the length of $\overline{DE}$ .
INSTRUCTION:
- Use the rounded answer from Question 1.
- Round your answer to two decimal places if necessary.
Answer: $\overline{DE} =$ $\text{cm}$
one-of
type(numeric.noerror)

ID is: 3558 Seed is: 8994

Algebra meets the sine rule

In triangle $ACB$ below, two angles are given: $\hat{A} = 60°$ and $\hat{B} = 87°$ . Side $\overline{BA}$ and $\overline{CB}$ have lengths $6 x - 2$ $\text{cm}$ and $4 x + 2$ $\text{cm}$ respectively, where $x \in \mathbb{R}$ . The third side, $\overline{AC}$ , is $\text{17}\ \text{cm}$ long. The figure may or may not be drawn to scale.

What is the value of $x$ ?

INSTRUCTION: Round your answer to two decimal places.

Answer: $x =$
one-of
type(numeric.abserror(0.004))
Hence determine the length of $\overline{BA}$ .
INSTRUCTION:
- Use the rounded answer from Question 1.
- Round your answer to two decimal places if necessary.
Answer: $\overline{BA} =$ $\text{cm}$
one-of
type(numeric.noerror)

ID is: 3571 Seed is: 7771

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides $z$ , $x$ , and $y$ , and angles $F$ , $H$ , and $G$ . Two of the three equal ratios are shown below the triangle.

\begin{matrix} \frac{y}{\sin H} = \frac{z}{\sin G} = third ratio \end{matrix}

$\begin{array}{c} \dfrac{ y }{ \sin H } = \dfrac{ z }{ \sin G } = \text{third ratio} \end{array}$

Which of the choices below is the third equal ratio for this triangle?

A	$\dfrac{ y }{ \sin F }$
B	$\dfrac{ x }{ \sin F }$
C	$\dfrac{ x }{ \sin H }$

Answer: The third ratio is choice .

ID is: 3571 Seed is: 148

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides $p$ , $m$ , and $n$ , and angles $H$ , $F$ , and $G$ . Two of the three equal ratios are shown below the triangle.

\begin{matrix} \frac{n}{\sin F} = \frac{p}{\sin G} = third ratio \end{matrix}

$\begin{array}{c} \dfrac{ n }{ \sin F } = \dfrac{ p }{ \sin G } = \text{third ratio} \end{array}$

Which of the choices below is the third equal ratio for this triangle?

A	$\dfrac{ p }{ \sin H }$
B	$\dfrac{ m }{ \sin G }$
C	$\dfrac{ m }{ \sin H }$

Answer: The third ratio is choice .

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The sine rule: the ambiguous case

In $\Delta LKJ$ , $\overline{LJ} = \text{60}$ and $\overline{JK} = \text{74}$ . The angle across from side $\overline{LJ}$ is $\hat{K} = \text{52,8}\text{°}$ . Calculate the size of the angle across from $\overline{JK}$ , $\hat{L}$ , if it is obtuse (greater than $\text{90}\text{°}$ ).

INSTRUCTION: Round your answers to two decimal places.

Answer:
The angle is $\text{°}$ .

numeric
When you use the sine rule, sometimes you get the ambiguous case, which leads to two solutions. But Question 1 was not ambiguous: it has only one solution. Why is this the case?

Answer:
Question 1 was not ambiguous because .

ID is: 3580 Seed is: 1853

The sine rule: the ambiguous case

In $\Delta MNP$ , $\overline{NM} = \text{49}$ and $\overline{PN} = \text{80}$ . The angle across from side $\overline{NM}$ is $\hat{P} = \text{37}\text{°}$ . Calculate the size of the angle across from $\overline{PN}$ , $\hat{M}$ , if it is obtuse (greater than $\text{90}\text{°}$ ).

INSTRUCTION: Round your answers to two decimal places.

Answer:
The angle is $\text{°}$ .

numeric
When you use the sine rule, sometimes you get the ambiguous case, which leads to two solutions. But Question 1 was not ambiguous: it has only one solution. Why is this the case?

Answer:
Question 1 was not ambiguous because .

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Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are $36°$ and $44°$ . The side across from the $36°$ angle is $\text{6}$ . Determine the length of the side across from the $44°$ angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is longer than the given side

6

$\text{6}$ . Check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: The length of the side is .

numeric

ID is: 3504 Seed is: 9479

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are $92°$ and $24°$ . The side across from the $92°$ angle is $\text{10}$ . Determine the length of the side across from the $24°$ angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is shorter than the given side

10

$\text{10}$ . Check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: The length of the side is .

numeric

ID is: 3551 Seed is: 2688

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides $\text{4}$ , $\text{5}$ , and $\text{7}$ , and two angles labelled $A$ and $B$ . You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

\begin{matrix} \frac{\sin B}{4} = \frac{\sin A}{7} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{4} = \dfrac{\sin A }{7} \end{array}$

What should you do to correct the equation?

Answer:

To correct the equation you should .

ID is: 3551 Seed is: 2685

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides $\text{7}$ , $\text{8}$ , and $\text{10}$ , and two angles labelled $A$ and $B$ . You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

\begin{matrix} \frac{10}{\sin B} = \frac{8}{\sin A} \end{matrix}

$\begin{array}{c} \dfrac{10}{\sin B } = \dfrac{8}{\sin A } \end{array}$

What should you do to correct the equation?

Answer:

To correct the equation you should .

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The sine rule: the ambiguous case

The diagram below shows $\Delta KJL$ . The diagram is not to scale. Side $LJ = \text{9,5}\ \text{cm}$ and side $JK = \text{13,4}\ \text{cm}$ . One angle is given: $\hat{K} = \text{43,6}\text{°}$ .

Determine the values for $\hat{L}$ . There are two correct answers.
INSTRUCTION:
- You can type your answers in any order.
- Round your answers to two decimal places.
Answer:
The possible answers are $\text{°}$ and $\text{°}$ .

numeric
numeric
If you are now told that $\hat{L}$ is obtuse, which of the answers to Question 1 is the correct answer?

Answer:
The correct answer must be $\text{°}$ .

numeric

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The sine rule: the ambiguous case

The diagram below shows $\Delta PMN$ . The diagram is not to scale. Side $MP = \text{9,5}\ \text{cm}$ and side $PN = \text{13,4}\ \text{cm}$ . One angle is given: $\hat{N} = \text{39,2}\text{°}$ .

Determine the values for $\hat{M}$ . There are two correct answers.
INSTRUCTION:
- You can type your answers in any order.
- Round your answers to two decimal places.
Answer:
The possible answers are $\text{°}$ and $\text{°}$ .

numeric
numeric
If you are now told that $\hat{M}$ is acute, which of the answers to Question 1 is the correct answer?

Answer:
The correct answer must be $\text{°}$ .

numeric

ID is: 1433 Seed is: 5382

ID is: 1433 Seed is: 1943

ID is: 3581 Seed is: 9755

Connecting the sine rule to right triangles

The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is $\text{90}\text{°}$ ?

Choices	If an angle is $\text{90}\text{°}$ , the sine rule...
A	will not work anymore.
B	becomes the same as the theorom of Pythagoras.
C	becomes the sine ratio (opposite & hypotenuse).
D	only uses the reciprocals.

Answer:

The correct statement is choice .

The table below shows (proves) that if we use the sine rule with a $\text{90}\text{°}$ angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

Answer:

Steps	Reasons
	Given
	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

ID is: 3581 Seed is: 1248

Connecting the sine rule to right triangles

The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is $\text{90}\text{°}$ ?

Choices	If an angle is $\text{90}\text{°}$ , the sine rule...
A	will give two answers due to the ambiguous case.
B	becomes the sine ratio (opposite & hypotenuse).
C	only works for the acute angles.
D	only uses the reciprocals.

Answer:

The correct statement is choice .

Answer:

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

ID is: 3500 Seed is: 8910

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. Two angles are shown, with values of $110°$ and $32°$ . The side of the triangle across from the $32^{ \circ }$ angle has a length of $\text{6,1}$ units.

Suppose you need to calculate $x$ , the side across from the $110^{ \circ }$ angle. The two equations below are both accurate and you can use either of these versions to find $x$ . But one choice makes the solution a bit easier. Which version is the better choice?

Version 1	$\dfrac{\sin \left(\text{110}°\right)}{x} = \dfrac{\sin \left(\text{32}° \right)}{ \text{6,1} }$
Version 2	$\dfrac{x}{\sin \left(\text{110}°\right)} = \dfrac{ \text{6,1} }{\sin \left(\text{32}° \right)}$

Answer: The better choice is: .

Now solve for the missing side, $x$ . Use the equation identified in Question 1.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The length of the missing side is units.
numeric

ID is: 3500 Seed is: 6618

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. Two angles are shown, with values of $87°$ and $38°$ . The side of the triangle across from the $87^{ \circ }$ angle has a length of $\text{13,5}$ units.

Suppose you need to calculate $x$ , the side across from the $38^{ \circ }$ angle. The two equations below are both accurate and you can use either of these versions to find $x$ . But one choice makes the solution a bit easier. Which version is the better choice?

Version 1	$\dfrac{\sin \left(\text{38}°\right)}{x} = \dfrac{\sin \left(\text{87}° \right)}{ \text{13,5} }$
Version 2	$\dfrac{x}{\sin \left(\text{38}°\right)} = \dfrac{ \text{13,5} }{\sin \left(\text{87}° \right)}$

Answer: The better choice is: .

Now solve for the missing side, $x$ . Use the equation identified in Question 1.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The length of the missing side is units.
numeric

ID is: 3556 Seed is: 8123

Working algebraically: finding a sine ratio

Triangle $DEF$ is shown below with sides labelled $12 y^{4}$ , $4 y$ , and $3 y - 4$ . All three of these expression are non-zero. $\angle E$ is labelled $\beta$ . The diagram is not drawn to scale.

Determine an expression for $\sin \beta$ in terms of $y$ and $\sin D$ .

Answer: $\sin \beta =$
expression
Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

Answer:
The expressions must be non-zero .

ID is: 3556 Seed is: 5355

Working algebraically: finding a sine ratio

Triangle $DEF$ is shown below with sides labelled $4 b - 3$ , $4 b$ , and $8 b^{5}$ . All three of these expression are non-zero. $\angle E$ is labelled $\theta$ . The diagram is not drawn to scale.

Determine an expression for $\sin \theta$ in terms of $b$ and $\sin D$ .

Answer: $\sin \theta =$
expression
Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

Answer:
The expressions must be non-zero .

Equation A	$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
Equation B	$\tan \theta = \dfrac{\sin \theta}{\cos \theta}$
Equation C	$a^{2} + b^{2} = c^{2}$
Equation D	$\tan \theta = m$

Steps	Reasons
	Given
$x^{2} + y^{2} = r^{2}$	Theorem of Pythagoras
$\dfrac{x^{2}}{r^{2}} + \dfrac{y^{2}}{r^{2}} = 1$	Multiply by $\dfrac{1}{r^{2}}$
	$\cos \theta = \dfrac{x}{r}$ on Cartesian plane
$\cos^{2} \theta + \sin^{2} \theta = 1$	$\sin \theta = \dfrac{y}{r}$ on Cartesian plane
$\sin^{2} \theta + \cos^{2} \theta = 1$

Equation A	$a^{2} + b^{2} = c^{2}$
Equation B	$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$
Equation C	$\tan \theta = m$
Equation D	$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

Steps	Reasons
	Given
$x^{2} + y^{2} = r^{2}$	Theorem of Pythagoras
$\dfrac{x^{2}}{r^{2}} + \dfrac{y^{2}}{r^{2}} = 1$
$\cos^{2} \theta + \dfrac{y^{2}}{r^{2}} = 1$	$\cos \theta = \dfrac{x}{r}$ on Cartesian plane
	$\sin \theta = \dfrac{y}{r}$ on Cartesian plane
$\sin^{2} \theta + \cos^{2} \theta = 1$	Commutativity of addition

Steps	Reasons
$4 \left(\sin \psi + \cos \psi\right) \left(\tan \psi \cdot \cos \psi - \cos \psi\right) + 8 \cos^{2} \psi$	Given
	Quotient identity substitution
$4 \left(\sin \psi + \cos \psi\right) \left(\sin \psi - \cos \psi\right) + 8 \cos^{2} \psi$
$4 \left(\sin^{2} \psi - \cos^{2} \psi\right) + 8 \cos^{2} \psi$	Multiply the brackets
$4 \sin^{2} \psi - 4 \cos^{2} \psi + 8 \cos^{2} \psi$	Distribute the $\text{4}$
$4 \sin^{2} \psi + 4 \cos^{2} \psi$
	Factorisation

$4$	Evaluate

Steps	Reasons
$4 \left(\sin \lambda + \cos \lambda\right) \left(\tan \lambda \cdot \cos \lambda - \cos \lambda\right) + 8 \cos^{2} \lambda$	Given
	Quotient identity substitution
$4 \left(\sin \lambda + \cos \lambda\right) \left(\sin \lambda - \cos \lambda\right) + 8 \cos^{2} \lambda$
$4 \left(\sin^{2} \lambda - \cos^{2} \lambda\right) + 8 \cos^{2} \lambda$	Multiply the brackets
	Distribute the $\text{4}$
$4 \sin^{2} \lambda + 4 \cos^{2} \lambda$
	Factorisation
$4 \left(1\right)$
$4$	Evaluate

Equation	Identity or not
$\left(2x\right)\left(2x^3\right) = 4x^4$
$x^2 - 2x - 3 = 0$
$5x + 4x = 9x$

Equation	Identity or not
$\left(4x\right)\left(5x^3\right) = 20x^4$
$2x + 2 = x + 3$
$4\left(x + 1\right) = 4x + 4$

Steps	Reasons
$\dfrac{ 8 }{ \cos^{2} A } - \dfrac{4 }{ 1 - \sin A}$	Given
	Multiply by $\dfrac{\left(1 + \sin A\right) }{ \left(1 + \sin A\right) } = 1$
$\dfrac{ 8 }{ \cos^{2} A } - \dfrac{4 \left(1 + \sin A\right) }{ \left(1 - \sin A\right) \left(1 + \sin A\right) }$
$\dfrac{ 8 }{ \cos^{2} A } - \dfrac{4 \left(1 + \sin A\right) }{ \left(1 - \sin^{2} A\right) }$	Expand brackets

$\dfrac{ 8 - 4 \left(1 + \sin A\right) }{ \left(1 - \sin^{2} A\right) }$	Add fractions with equal denominators
$\dfrac{ 8 - 4 - 4 \sin A }{ \left(1 - \sin^{2} A\right) }$	Distribute

$\dfrac{ 4 - 4 \sin A }{ \left(1 + \sin A\right) \left(1 - \sin A\right) }$	Factorise difference of two squares
$\dfrac{ 4 \left(1 - \sin A\right) }{ \left(1 + \sin A\right) \left(1 - \sin A\right) }$	Factorise out $\text{4}$
$\dfrac{ 4 }{ 1 + \sin A }$	Cancel $\left(1 - \sin A\right)$ factors

Steps	Reasons
$\dfrac{ 12 }{ \sin^{2} B } - \dfrac{6 }{ 1 - \cos B}$	Given
$\dfrac{ 12 }{ \sin^{2} B } - \dfrac{6 }{ \left(1 - \cos B\right)} \cdot \dfrac{ \left(1 + \cos B\right) }{ \left(1 + \cos B\right) }$
$\dfrac{ 12 }{ \sin^{2} B } - \dfrac{6 \left(1 + \cos B\right)}{ \left(1 - \cos B\right) \left(1 + \cos B\right) }$	Multiply fractions

$\dfrac{ 12 }{ \left(1 - \cos^{2} B\right) } - \dfrac{6 \left(1 + \cos B\right) }{ \left(1 - \cos^{2} B\right)}$	Square identity substitution
$\dfrac{ 12 - 6 \left(1 + \cos B\right) }{ \left(1 - \cos^{2} B\right) }$	Add fractions with equal denominators
$\dfrac{ 12 - 6 - 6 \cos B }{ \left(1 - \cos^{2} B\right) }$	Distribute
	Collect constant terms
$\dfrac{ 6 - 6 \cos B }{ \left(1 + \cos B\right) \left(1 - \cos B\right) }$	Factorise difference of two squares

$\dfrac{ 6 }{ 1 + \cos B }$	Cancel $\left(1 - \cos B\right)$ factors

	An identity is an equation that...
A	is true for all values of $\theta$ for which both sides are defined.
B	is false when there is an undefined value
C	is true for all values of $\theta$ which are inside the interval.
D	is true unless it has no solution.

Steps	Reasons
$-\dfrac{ 3 \left(1 - \cos \theta \right) \left(1 + \cos \theta \right) }{ 5 \sin^{2} \theta }$
$-\dfrac{ 3 \left(1 - \cos^{2} \theta\right) }{ 5 \sin^{2} \theta }$
	Square identity substitution
	Cancel $\sin \theta$ factors

Steps	Reasons
$\dfrac{\tan \theta - \tan \theta \cdot \sin^{2} \theta }{ \cos^{2} \theta }$
$\dfrac{\tan \theta \left(1 - \sin^{2} \theta\right) }{ \cos^{2} \theta }$
	Square identity substitution
	Cancel cosine factors

1	$\left(1 - \cos A\right)\left(1 + \cos A\right)$
2	$\left(1 + \cos A\right)\left(1 + \cos A\right)$
3	$\left(1 - \sin A\right)\left(1 + \sin A\right)$
4	$1 - \sin^{2} A$

1	$1 - \cos^{2} \beta$
2	$1 - \sin^{2} \beta$
3	$\left(1 - \cos \beta\right)\left(1 + \cos \beta\right)$
4	$\left(1 - \sin \beta\right)\left(1 - \sin \beta\right)$

A	$\dfrac{\sin \alpha}{\cos \alpha}$
B	$\sin^{2} \alpha + \cos^{2} \alpha$
C	$1$

A	$\dfrac{\sin \alpha}{\cos \alpha}$
B	$\sin^{2} \alpha + \cos^{2} \alpha$
C	$1$

A	$\tan\left ( 180°- x\right ) = + \tan\left (x\right )$
B	$\tan\left ( 180°- x\right ) = - \tan\left (x\right )$

A	$\cos\left ( 180°- y\right ) = + \cos\left (y\right )$
B	$\cos\left ( 180°- y\right ) = - \cos\left (y\right )$

Steps	Reasons
$\left(\tan \theta + 1\right) \cdot \left(\tan \theta - 1\right) \cdot \cos^{2} \theta$	Given
$\left(\tan^{2} \theta - 1\right) \cdot \cos^{2} \theta$
$\tan^{2} \theta \cdot \cos^{2} \theta - \cos^{2} \theta$
$\left(\dfrac{ \sin^{2} \theta } { \cos^{2} \theta }\right) \cdot \cos^{2} \theta - \cos^{2} \theta$	Quotient identity substitution
	Cancel $\cos^{2} \theta$ factors
$\left(1 - \cos^{2} \theta\right) - \cos^{2} \theta$	Square identity substitution
	Collect $\cos^{2} \theta$ terms

Steps	Reasons
$\dfrac{ 3 \sin \theta \cdot \cos \theta } { \tan \theta } + 3 \sin^{2} \theta$	Given
$\dfrac{ 3 \sin \theta \cdot \cos \theta } { \left(\frac{ \sin \theta } { \cos \theta }\right) } + 3 \sin^{2} \theta$
$3 \sin \theta \cdot \cos \theta \cdot \left(\dfrac{ \cos \theta }{ \sin \theta }\right) + 3 \sin^{2} \theta$	Change division to multiplication
$3 \cos \theta \cdot \cos \theta + 3 \sin^{2} \theta$	Cancel sine factors
	Combine cosine factors
	Factorisation
$3 \left(\sin^{2} \theta + \cos^{2} \theta\right)$	Commutativity of addition
$3 \left(1\right)$	Square identity substitution
$3$

Equation A	$\tan x = \dfrac{\sin x}{\cos x}$
Equation B	$\sin ^{2} x + \cos ^{2} x = 1$
Equation C	$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

Steps	Reasons
$\dfrac{\sin \theta }{ 1 - \cos \theta }$
$\dfrac{\sin \theta }{ \left(1 - \cos \theta\right) } \cdot \dfrac{ \left(1 + \cos \theta\right)}{ \left(1 + \cos \theta\right) }$	Multiply by $\dfrac{ \left(1 + \cos \theta\right) }{\left(1 + \cos \theta\right)}=1$
$\dfrac{\sin \theta \left(1 + \cos \theta\right) }{ \left(1 - \cos \theta\right) \left(1 + \cos \theta\right) }$
$\dfrac{\sin \theta \left(1 + \cos \theta\right) }{ \left(1 - \cos^{2} \theta\right) }$	Expand brackets
	Square identity substitution
	Cancel $\sin \theta$ factors

Equation A	$\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$
Equation B	$\tan \theta = m$
Equation C	$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$
Equation D	$\cos^{2} \theta = 1 - \sin^{2} \theta$

Steps	Reasons
$\tan \theta = \dfrac{\text{opp}}{\text{adj}}$	Definition of tangent ratio
$\tan \theta = \dfrac{\text{opp}}{\text{adj}} \cdot \dfrac{\frac{1}{\text{hyp}}}{\frac{1}{\text{hyp}}}$
$\tan \theta = \dfrac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{adj}}{\text{hyp}}}$	Fraction multiplication
	Definitions of sine and cosine ratios

A	$a < x < b$
B	$a \leq x \leq b$
C	$-\text{180}\text{°} \leq x < a \:\: \text{or} \:\: b < x \leq \text{0}\text{°}$
D	$-\text{180}\text{°} \leq x \leq a \:\: \text{or} \:\: b \leq x \leq \text{0}\text{°}$

A	$a < x < b$
B	$a \leq x \leq b$
C	$-\text{120}\text{°} \leq x < a \:\: \text{or} \:\: b < x \leq \text{0}\text{°}$
D	$-\text{120}\text{°} \leq x \leq a \:\: \text{or} \:\: b \leq x \leq \text{0}\text{°}$

Choices	If you use a $90°$ angle is the cosine rule...
A	it becomes the theorem of Pythagoras
B	it changes into the sine rule
C	it becomes the equilateral triangle calculation
D	the sum of the interior angles of a triangle is $180°$

Proof Steps
1.	$= \dfrac{x}{a}$
2.	$\therefore x = a\cos{\hat{C}}$
	In $\Delta BAD$
3.	$c^{2} \;=\;$ $\;+\; (b - x)^{2} \;$ $(\text{Pythagoras})$
4.	$\therefore BD^{2} = \;$ $c^{2} - b^{2} + 2bx - x^{2}$
	In $\Delta BDC$
5.	$a^{2} = BD^{2} + x^{2} \;$ $(\text{Pythagoras})$
6.	$\therefore BD^{2} =$
7.	$c^{2} - b^{2} + 2bx - x^{2}$ $\;=\;$ $a^{2} - x^{2}$
8.	$c^{2} = b^{2} + a^{2} +$
9.	$\therefore c^{2} =$ $\;a^{2} + b^{2}$ $\; -\; 2ab\cos{\hat{C}}$

Trigonometry

1. Revision

CAST diagram: problem solving to find coordinates

CAST diagram: problem solving to find coordinates

2. Trigonometric identities

Where does the square identity come from?

Where does the square identity come from?

Using the square identity to simplify an expression

Using the square identity to simplify an expression

Simplifying an identity using tan𝐴=sin𝐴cos𝐴tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}

Simplifying an identity using tan𝐴=sin𝐴cos𝐴tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}

Proofs: trigonometric identities

Proofs: trigonometric identities

Identity or not?

Identity or not?

Proofs: trigonometric identities

Proofs: trigonometric identities

Evaluate an expression using trig identities

Evaluate an expression using trig identities

Asymptotes of the tangent function

Asymptotes of the tangent function

Undefined cases

Undefined cases

Simplifying trigonometric identities

Simplifying trigonometric identities

Simplifying an identity using sin2𝐴+cos2𝐴=1sin2⁡A+cos2⁡A=1\sin^{2} A + \cos^{2} A = 1

Simplifying an identity using sin2𝐴+cos2𝐴=1sin2⁡A+cos2⁡A=1\sin^{2} A + \cos^{2} A = 1

Simplifying trigonometric identities

Simplifying trigonometric identities

Different forms of the quotient identity

Different forms of the quotient identity

Building a proof

Building a proof

When is an identity undefined?

When is an identity undefined?

Finding values for which an identity is undefined

Finding values for which an identity is undefined

Different forms of the square (Pythagorean) identity.

Different forms of the square (Pythagorean) identity.

Proofs: trigonometric identities

Proofs: trigonometric identities

Naming important identities

Naming important identities

Properties of a proof

Properties of a proof

Proving an identity

Proving an identity

Where does the quotient identity come from?

Where does the quotient identity come from?

Definition: identity

Definition: identity

Using the quotient identity to simplify an expression

Using the quotient identity to simplify an expression

Proving trig identities: difference of two squares

Proving trig identities: difference of two squares

Examining identities

Examining identities

Remembering the trigonometric identities

Remembering the trigonometric identities

3. Reduction formula

Simplifying trigonometric expressions with reduction formulas

Simplifying trigonometric expressions with reduction formulas

Applying reduction formulas

Applying reduction formulas

Applying reduction formulas

Applying reduction formulas

Reduction formulas for the trigonometric functions

Reduction formulas for the trigonometric functions

Simplifying trigonometric expressions with reduction formulas

Simplifying trigonometric expressions with reduction formulas

Pythagorean problems

Pythagorean problems

Reduction formulas: "out of order" angles with negatives

Reduction formulas: "out of order" angles with negatives

Applying reduction formulas

Applying reduction formulas

Trigonometric expressions

Trigonometric expressions

Trig: Pythagorean problems in the Cartesian plane

Trig: Pythagorean problems in the Cartesian plane

Simplifying an identity using $\tan A = \frac{\sin A}{\cos A}$

Simplifying an identity using $\tan A = \frac{\sin A}{\cos A}$

Simplifying an identity using $\sin^{2} A + \cos^{2} A = 1$

Simplifying an identity using $\sin^{2} A + \cos^{2} A = 1$

Proof Steps
1.	$A(b; 0) \;$ $B($ $; a\sin{\hat{C}}) \;$ $C(0; 0)$
2.	$AB^{2} = AD^{2} + BD^{2} \;$ $(\text{Pythagoras})$
3.	$c^{2} =$ $\; (b - a\cos{\hat{C}})^{2} \;+\;$
4.	$c^{2} =$ $\; b^{2} - 2ab\cos{\hat{C}} + a^{2}\cos^{2}{\hat{C}} \;+\;$
5.	$c^{2} =$ $\; b^{2} - 2ab\cos{\hat{C}}$ $\;+\; a^{2}(\cos^{2}{\hat{C}} + \sin^{2}{\hat{C}})$
6.	$c^{2} =$ $\;b^{2}$ $\;-\; 2ab\cos{\hat{C}}$ $\;+\; \;$ $(1)$
7.	$\therefore c^{2} =$ $\;a^{2} + b^{2}$ $\; -\; 2ab\cos{\hat{C}}$